Lebesuge integrable function always bounded?

Question

Assume $f: [0,1] \to \mathbb{R}$ is Lebesgue integrable, does it imply that $f$ is bounded almost surly?

$\frac{1}{2\sqrt{x}}$ is not bounded but its integrable: $\int_0^1\frac{1}{2\sqrt{x}}, dx=1$ — Bman72, Aug 27 '14 at 10:20

score 1 · Accepted Answer · answered Aug 27 '14 at 09:57

1

I don't think so..

Look for example at $\frac1{\sqrt{x(1-x)}}$.

answered Aug 27 '14 at 09:57

D.L.

1 Answers1